STAT 400 Lecture AL1 Answers for 1.5 Spring 2015 Dalpiaz 1. In Neverland, men constitute 60% of the labor force. The rates of unemployment are 6.0% and 4.5% among males and females, respectively. A person is selected at random from Neverland’s labor force. P( M ) = 0.60, P( U | M ) = 0.06, P( U | F ) = 0.045. a) What is the probability that the person selected is a male and is unemployed? P( M ∩ U ) = P( M ) × P( U | M ) = 0.60 × 0.06 = 0.036. b) What is the probability that the person selected is a female and is unemployed? P( F ∩ U ) = P( F ) × P( U | F ) = 0.40 × 0.045 = 0.018. Unemployed Employed Total Male 0.036 0.564 0.60 Female 0.018 0.382 0.40 Total . 0.054 0.946 1.00 c) What is the probability that the person selected is unemployed? P( U ) = 0.036 + 0.018 = 0.054. OR Law of Total Probability: P( U ) = P( M ) × P( U | M ) + P( F ) × P( U | F ) = 0.60 × 0.06 + 0.40 × 0.045 = 0.054. d) Suppose the person selected is unemployed. What is the probability that a male was selected? P( M | U ) = 0.036/0.054 = 2/3. OR Bayes’ Theorem: P( M | U ) = ) F| UP() F P() M| UP( ) M P() M| UP( ) M P(× + ×× = 045.040.006.060.006.060.0×+×× = 32.2. In a presidential race in Neverland, the incumbent Democrat ( D ) is running against a field of four Republicans ( R 1 , R 2 , R 3 , R 4 ) seeking the nomination. Political pundits estimate that the probabilities of R 1 , R 2 , R 3 , and R 4 winning the nomination are 0.40, 0.30, 0.20, and 0.10, respectively. Furthermore, results from a variety of polls are suggesting that D would have a 55% chance of defeating R 1 in the general election, a 70% chance of defeating R 2 , a 60% chance of defeating R 3 , and an 80% chance of defeating R 4 . Assuming all these estimates to be accurate, what are the chances that D will be a two-term president? P( R 1 ) = 0.40, P( R 2 ) = 0.30, P( R 3 ) = 0.20, P( R 4 ) = 0.10, P( W | R 1 ) = 0.55, P( W | R 2 ) = 0.70, P( W | R 3 ) = 0.60, P( W | R 4 ) = 0.80. Law of Total Probability: P( W ) = P( W ∩ R 1 ) + P( W ∩ R 2 ) + P( W ∩ R 3 ) + P( W ∩ R 4 ) = P( R 1 ) P( W | R 1 ) + P( R 2 ) P( W | R 2 ) + P( R 3 ) P( W | R 3 ) + P( R 4 ) P( W | R 4 ) = 0.40 ⋅ 0.55 + 0.30 ⋅ 0.70 + 0.20 ⋅ 0.60 + 0.10 ⋅ 0.80 = 0.63. R 1 R 2 R 3 R 4 W 0.40 ⋅ 0.55 0.22 0.30 ⋅ 0.70 0.21 0.20 ⋅ 0.60 0.12 0.10 ⋅ 0.80 0.08 0.63 L 0.18 0.09 0.08 0.02 0.37 0.40 0.30 0.20 0.10 1.003. In Anytown, 10% of the people leave their keys in the ignition of their cars. Anytown’s police records indicate that 4.2% of the cars with keys left in the ignition are stolen. On the other hand, only 0.2% of the cars without keys left in the ignition are stolen. Suppose a car in Anytown is stolen. What is the probability that the keys were left in the ignition? P( Keys ) = 0.10, P( Keys' ) = 1 − 0.10 = 0.90. P( Stolen Keys ) = 0.042. P( Stolen Keys' ) = 0.002. Need P( Keys Stolen ) = ? Bayes’ Theorem: P(Keys Stolen) = )KeysP(Stolen)P(KeysKeys)P(StolenP(Keys)Keys)P(StolenP(Keys)|||''×+×× = 002.090.0042.010.0042.010.0×+×× = 0.70. OR Stolen' Stolen 0.042 ⋅ 0.10 Keys 0.0042 0.0958 0.10 Keys' 0.002 ⋅ 0.90 0.0018 0.8982 0.90 0.0060 0.9940 1.00 P( Keys Stolen ) = 0060000420)Stolen P()Stolen Keys P(..=∩ = 0.70. ORP( Stolen ) = 0.0042 + 0.0018 = 0.0060. P( Keys Stolen ) = 0060000420)Stolen P()Stolen Keys P(..=∩ = 0.70.3 ¼ . A warehouse receives widgets from three different manufacturers, A (50%), B (30%), and C (20%). Suppose that 2% of the widgets coming from A are defective, as are 4% of the widgets coming from B, and 7% coming from C. a) Find the probability that a widget selected at random at this warehouse is defective. Law of Total Probability: P( D ) = P( A ) × P( D | A ) + P( B ) × P( D | B ) + P( C ) × P( D | C ) = 0.50 × 0.02 + 0.30 × 0.04 + 0.20 × 0.07 = 0.010 + 0.012 + 0.014 = 0.036. b) Suppose a widget that was selected at random is found to be defective. What is the probability that it came from manufacturer A? Manufacturer B? Manufacturer C? P( A | D ) = 036.0010.0 = 185. P( B | D ) = 036.0012.0 = 186. P( C | D ) = 036.0014.0 = 187.3 ½ . Seventy percent of the light aircraft that disappear while in flight in Neverland are subsequently discovered. Of the aircraft that are discovered, 60% have an emergency locator, whereas 90% of the aircraft not discovered do not have such a locator. Suppose a light aircraft that has just disappeared has an emergency locator. What is the probability that it will not be discovered? P ( Discovered ) = 0.70, P ( Discovered ' ) = 1 − 0.70 = 0.30. P ( Locator Discovered ) = 0.60. P ( Locator ' Discovered ' ) = 0.90. Need P ( Discovered ' Locator ) = ? Locator ' Locator 0.60 ⋅ 0.70 Discovered 0.42 0.28 0.70 Discovered ' 0.90 ⋅ 0.30 0.03 0.27 0.30 0.45 0.55 1.00 P ( Discovered ' Locator ) = 151==∩450030)Locator P()Locator Discovered P( ..' ≈ 0.06667.4. In a certain population, the proportion of individuals who have a particular disease is 0.025. A test for the disease is positive in 94% of the people who have the disease and in 4% of the people who do not. P( D ) = 0.025, P( + | D ) = 0.94, P( + | D' ) = 0.04. a) Find the probability of receiving a positive reaction from this test. Need P( + ) = ? + – D 0.025 ⋅ 0.94 0.0235 0.0015 0.025 D' 0.975 ⋅ 0.04 0.0390 0.9360 0.975 0.0625 0.9375 1.000 OR Law of Total Probability: P( + ) = P( D ) × P( + | D ) + P( D' ) × P( + | D' ) = 0.025 × 0.94 + 0.975 × 0.04 = …
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